Balance the following equation using oxidation number method
`Ag_(2)S_(3) +HNO_(3) +H_(2)O to H_(3)AsO_(4) +H_(2)SO_(4) +NO`

Explanation

Equate the total no. of electrons in the reactant side by cross multiplying,
`rArr 3As_(2)S_(3) +28HNO_(3) + H_(2)O to H_(3)AsO_(4) +H_(2)SO_(4) +NO`
Based on reactant side, balance the products
`rArr 3As_(2)S_(3) +28HNO_(3) +H_(2)O to 6H_(3)AsO_(4) +9H_(2)SO_(4) +28NO`
Product side :36 hydrogen atoms & 88 oxygen atoms
Reactant side :28 hydrogen atoms & 74 oxygen atoms
Difference is 8 hydrogen atoms & 4 oxygen atoms
`:.` Multiply `H_(2)O` molecule on the reactant side by .4..
Balanced equation is,
`3As_(2)S_(3) +28HNO_(3) +4H_(2)O to 6H_(3)AsO_(4) +9H_(2)SO_(4) +28NO`